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Chapter 11 Gear Trains
§11－1 Gear Trains and Their Classification
§11－2 Train Ratio of a Gear Train With Fixed
Axes
§11－3 Train Ratio of Elementary Epicyclic
Gear Train
§11－4 Train Ratio of a Combined Gear Train
§11－5 Applications of Gear Trains
§11－1 Gear Trains and Their Classification
3
Gear Train——A transmission system by
2
3′
more than one pair of gears.
1
Classification of Gear Trains
4
一、 Gear train with fixed axes
4′
(Ordinary gear train)
5
Definition—The positions of all gear axes
in a gear train are fixed
O1
H
二、 Epicyclic gear train
2
O1
2
O
O
Definition—At least one gear axis
in a gear train rotates
the other axis.
3
H
1
3
1
Differential gear
train: F=2
三、Combined gear train
Definition—is a combination
of several elementary epicyclic
gear trains or a combination
of at least one elementary
epicyclic gear train with at
least one gear train with fixed
axes.
O1
H
2
O1
3
2
O
O
1
H
1
3
planetary gear train: F=1
§11－2 Train Ratio of a Gear Train With Fixed Axes
The train ratio—the ratio of the angular velocities of input and
output members in the gear train.
3
Train ratio includes two factors,
1) magnitude and
2)relative rotating direction of the two members.
2
3′
1
一、Calculations of train ratio
Given:the number of all gears，
Find train ratio i15 = 1 / 5 。
z2 z3 z4 z5
1
i15 
 i12  i2 3  i3' 4  i4 5 
5
z1 z2 z3' z4
Train ratio of gear
train with fixed axes
Product of tooth numbers of all the driven gears
＝
Product of tooth numbers of all the driving gears
4
4′
5
1
z
 2
2
z1
z

 2  3
3
z2
i12 
i23
3
z
 4
4
z3
z

 4  5
5
z4'
i3' 4 
i4' 5
二、Relative rotating directions of gears
The rotating direction of the input gear 1 is given as shown in
Fig. Where the direction of the arrow indicates that of the
peripheral velocity of the gear on the visible side.
1） The axis of the input gear is parallel to the axis of the output gear
The directions of the input gear and the output
gear are the same，train ratio is“+”; the
3
2
3′
directions are opposite, train ratio is“-”.
2） The axis of the input gear is not parallel to the
axis of the output gear
It can only be determined by drawing arrows.
z2 z3 z4 z5
1
i15 

5
z1 z2 z3' z4
1
4
4′
5
idle wheel——The function of gear 2 is to change the rotating
direction of the output gear , not the magnitude of the train ratio.
§11－3 Train Ratio of Elementary Epicyclic Gear Train
Shown in a is a typical elementary epicyclic gear train. It consists of two
sun gears , one planet gear , one planet carrier H, and the frame. Suppose that
the planet carrier H rotates in an angular velocity wH. The train ratio of this
gear train cannot be calculated simply as a gear train with fixed axes because of
the rotation of the planet carrier.
-ωH
Imagine that we add an
angular velocity (- wH) to the whole
O
gear train. It will keep the relative
motion between any two links
unchanged. Now the angular velocity
of the planet carrier H is wH - wH =0.
This makes the planet carrier H
fixed, as shown in Fig.
O1
H
2
O1
3
2
O
1
ω3
ω2
ωH H
1 ω1
3
To add a common angular velocity “-ωH” to the whole epicyclic
gear train, the changes of angular velocity of each component
are shown in the table below:
Components
The former angular
Angular velocities in the converted
velocities
gear trains
Gear 1
ω1
1H  1  H
Gear 2
ω2
2H  2  H
Gear 3
ω3
3H  3  H
H
ωH
HH  H  H  0
Since the converted gear train is a gear train with fixed axes,
its train ratio can be calculated as for an ordinary gear train.
H

1  H
H
1
i13  H 
3 3  H
z2 z3
z3


z1 z2
z1
O1
H
2
O1
3
2
H
O
O
1
3
1
Where the“－” sign demonstrates that gears 1 and 3 in the
converted gear train rotate oppositely.
H

m   H
H
m
imn  H 
n n   H
product of tooth numbers of all the driven gears in the converted gear train

product of tooth numbers of all the driving gears in the converted gear train
When using the equation, please notice the following：
1）The equation above are only fit for the case that axes of the sun
gear m, n and the planet carrier H are parallel.
2）When determining the train ratio of an epicyclic gear train,we
enter the values along with the signs of angular velocities of ωm,ωn
andωH for the equation above. If we designate some direction of
rotation as the positive one, then the opposite direction as negative.
3）Judging the “+” “-” signs, we consider the planet carrier H as
stationary, then the method same to the ordinary gear trains can be
used.
As regards planetary gear trains, that is ωn＝0, the Eq. can be
rewritten as：
H
mn
i
m   H

 imH  1
0  H
imH  1  i
H
mn
Example11-1 In the gear train in Fig., z1 = z2 = 30, z3＝90, n1 = 1
r/min counterclockwise, n3 = 1 r/min clockwise. (the value of angular
velocity is defined as positive when counterclockwise). Find nH.
Solution：
z2 z3
z3
n1  nH
i 


n3  nH
z1 z2
z1
z3
1  nH
90
     3
1  nH
z1
30
H
13
then
nH= -1/2
O1
H
2
O1
O
O
i1H = n1 / nH = -2
3
1
Example 11-2 In the planetary gear train shown in figure,
z1 =100, z2 = 101, z2’=100, z3= 99. Find the train ratio iH1.
Solution:
2
2’
For the planetary gear train
shown in the figure,
 z 2 z3 
i1H  1  i  1   

 z2 z2' 
1
 101 99 
 1 

 100  100  10000
H
13
Therefore,
iH1= 1 / i1H = 10000
H
1
3
§11－4 Train Ratio of a Combined Gear Train
z1= 20
z3= 30
Calculating the train ratio of a combined gear train：
z4= 80
1.To divide the combined gear train into several elementary
epicyclic gear trains and ordinary gear trains；
z’2= 20
2. to write their train ratio equations independently.
Solution：planet gears
z2= 40
planet carriers
sun gears
Example: Shown in the gear train, the numbers of all gears are given.Calculate
the train ratio i1H .
Solution
ordinary gear trains：
Elementary epicyclic gear trains：
gears 1、2

z
i12  1   2
2
z1

40
 2
20
i1H  i12i2' H  2  5  10
gears 3、H、 2’、and 4
i2' H  1  i
H
2'4
 1
 z4 
 1  

z

2' 
80
5
20
§11－5 Applications of Gear Trains
1. Branching transmission
2. To get a large train ratio
3. To change the speed of rotation
4. To change the direction of rotation
5. To combine the motion
6. To resolve the motion
7. To get a large power transmission
Fig. shows differential
mechanism of car，
where： Z1= Z3 ，nH= n4
n1  nH
z3
=－1
i 

n3  nH
z1
H
13
nH 
1
n1  n3 
2
The motor drives gear 5 through the transmission shaft.
When a car move forward straight,
n1 =n3
P
r
v3
v1
ω
When a car turns left, the car rotates around the point P：
r－radius
V1=(r-L) ω
2L－distance between two wheels
V3=(r+L) ω
n1 /n3 = V1 / V3 = (r-L) / (r+L)
rL
n1 
nH
r
2L
n3 
rL
nH
r
move straight turn corner
5
2
4
1
3
2
H
Differential mechanism